Airplane Velocity: 151.97 km/h NW

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An airplane traveling due west at 150.0 km/h experiences a northward wind of 24.4 km/h, resulting in a ground speed of approximately 151.97 km/h. The angle of the airplane's velocity relative to the ground is calculated using trigonometry, yielding an angle of 9.23 degrees north of west. This angle is measured from the westward direction, aligning with the mathematical convention of angles. The discussion highlights the distinction between mathematical angle measurements and compass directions, clarifying that the angle is indeed in the fourth quadrant. Understanding these concepts is crucial for accurately interpreting the airplane's velocity relative to the ground.
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An airplane is heading due west. The airplane's speed relative to the air is 150.0 km/h. If there is a wind of 24.4km/h toward the north, what is a) the magnitude (km/h)

I used the equation sqrt 150.0^2+24.4^2 =151.97

b)what is the direction (degrees measured from west) of the velocity of the plane relative to the ground?

tan-1(24.4/150.0)=9.23 Would you add 90degrees to this =99.23
 
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Well since the wind is north and the plane is heading due west, it must be within 270 and 360 degrees (on a compass).
 
I don't understand that.

Since it is heading due west, which would be on the -x axis. And the wind is blowing north wouldn't the angle be in the 2nd quadrant relative to the ground. Which would place the vector in the second quadrant between 90 and 180 degrees?

Can you elaborate on your previous response?
 
The question asks for "degrees measured from west". I would answet it as "9.23 degrees north of west"

Compass courses differ from the angles that are used in mathematics N 0, E 90, S 180, W 270.
 
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